Question:

<p><strong>Step 1: Simplify the first fraction</strong></p><p>Note that x + 1 = (x^(1/3))³ + 1³, which factors as a sum of cubes:</p><p>x + 1 = (x^(1/3) + 1)(x^(2/3) - x^(1/3) + 1)</p><p>Therefore: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3} + 1$</p><p><strong>Step 2: Simplify the second fraction</strong></p><p>Factor x - 1 and x - x^(1/2):</p><p>$\frac{x-1}{x-x^{1/2}} = \frac{(x^{1/2})^2 - 1}{x^{1/2}(x^{1/2}-1)} = \frac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)} = \frac{x^{1/2}+1}{x^{1/2}} = 1 + x^{-1/2}$</p><p><strong>Step 3: Combine the simplified expressions</strong></p><p>$\left(x^{1/3}+1-(1+x^{-1/2})\right)^{10} = \left(x^{1/3}-x^{-1/2}\right)^{10}$</p><p><strong>Step 4: Find the independent term</strong></p><p>General term: $\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r = \binom{10}{r}(-1)^r x^{(10-r)/3-r/2}$</p><p>For independent term: $\frac{10-r}{3} - \frac{r}{2} = 0$</p><p>$\frac{2(10-r)-3r}{6} = 0 \Rightarrow 20-5r = 0 \Rightarrow r = 4$</p><p><strong>Step 5: Calculate the coefficient</strong></p><p>$\binom{10}{4}(-1)^4 = \frac{10!}{4!6!} = 210$</p><p>∴ Answer: A (210)</p>